Question

9.3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution
are:
(a) 6.68 x 10^23(b) 6.09 x 10^22(c) 6.022 x 10^29(d) 6.022 x 10^21​

Answer 1

Answer:

Explanation:

molar mass of sucrose (C12H22O11) = 342 grams

so 1 mole of sucrose = 342 grams

moles = mass given/ molar mass

moles= 3.42/342=0.01 moles

1 mole of sucrose or 342 gram of sucrose contains oxygen atoms = 11 mole of Oxy. = 11*6.022*10^23 atoms of Oxy.

so by unitary method, 3.42  gram  or 0.01 mole of sucrose contains oxygen atom =   (11*6.022*10^3*0.01) = 6.62*10^22 atoms ( eq. 1)

18 gram of water (h2o)  contains oxygen atoms = 6.022*10^23 atoms (eq.2)

total no. of oxygen atom in solution = 6.62*10^22 +6.022*10^23

=6.62*10^22+60.22*10^22 = 66.84*10^22 atoms of oxygen or 6.68*10^23 atoms of oxygen in solution.

option A

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